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NCERT Class X Chapter 4:Quadratic Equation Exercise 4.2 Question 2 Question: Solve the problems given in Example 1. Solution: Example 1 subdivision i solution Example 1 subdivision ii solution Next Question Solution: NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 3. Explore more in Quadratic Equations chapter: Click this link to explore more NCERT Class X Chapter 4 Quadratic Equations solutions Explore more: Click this link to explore more NCERT Class X chapter solutions © Kaliyuga Ekalavya. All rights reserved.

NCERT Class X Chapter 4:Quadratic Equation Exercise 4.2 Question 3 Question: Find two numbers whose sum is 27 and product is 182. Given: Sum of two numbers = 27 Product of two numbers = 182 To Find: The two numbers. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: Let one of the numbers be 'x' and the other number be 'y' ⇒x + y = 27 ⇒y = 27 - x According to the given condition, their product is ...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 4 Question: Find two consecutive positive integers, sum of whose squares is 365. Given: Two consecutive positive integers. Sum of their squares = 365. To Find: The two consecutive positive integers. Formula: WKT, if the first integer is x, the next consecutive integer is (x + 1). WKT, the algebraic identity (a + b) 2 = a 2 + 2ab + b 2 . WKT, For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs ....

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 5 Question: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Given: Type of triangle: Right triangle. Altitude = Base - 7 cm. Hypotenuse = 13 cm. To Find: The lengths of the base and altitude. Formula: WKT, for a right triangle, by Pythagorean Theorem: Base 2 + Altitude 2 = Hypotenuse 2 . WKT, For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m ...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (iv) Question: Find the roots of the following quadratic equations by factorisation : 2x 2 – x + (1/8) = 0 Given: The quadratic equation: 2x 2 – x + 1 8 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 2x 2 – x + 1 8 = 0. First, clear the fra...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 6 Question: A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article. Given: Cost of production of each article = 3 + 2 × (Number of articles produced). Total cost of production = Rs 90. To Find: The number of articles produced and the cost of each article. Formula: WKT, Total Cost of Production = Number of Articles Produced × Cost of Production of Each Article. WKT, For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                          ...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (v) Question: Find the roots of the following quadratic equations by factorisation : 100x 2 – 20x + 1 = 0 Given: The quadratic equation: 100x 2 – 20x + 1 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 100x 2 – 20x + 1 = 0. Here, a = 100, b = -20, c = 1. Product (a × c)...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (iii) Question: Find the roots of the following quadratic equations by factorisation: √2x 2 + 7x + 5√2 = 0 Given: The quadratic equation: √2x 2 + 7x + 5√2 = 0 To Find: The values of x (roots) for the given quadratic equation. Formula: WKT, For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given quadratic equation is √2x 2 + 7x + 5√2 = 0. Comparing the given quadr...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (ii) Question: Find the roots of the quadratic equation:Find the roots of the quadratic equation 2x 2 + x – 6 = 0 . Given: The quadratic equation: 2x 2 + x – 6 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 2x 2 + x – 6 = 0. Here, a = 2, b = 1, c = -6. Product (a × c) ...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (i) Question: Find the roots of the following quadratic equations by factorisation: x 2 – 3x – 10 = 0 Given: The quadratic equation: x 2 – 3x – 10 = 0 To Find: The roots of the equation by factorization. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is x 2 – 3x – 10 = 0. Here, a = 1, b = -3, c = -10. Product (a × c) = 1 × (-1...